Tag: motional emf

Questions Related to motional emf

A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of 40 rad $s^{-1}$ in a uniform horizontal magnetic field of magnitude $2 \times 10^{-2}$ T. If the coil form a closed loop of resistance 8 $\Omega$, then the average power loss

motional emf physics

  1. $2.07\times 10^{-3}$W

  2. $1.23\times 10^{-3}$W

  3. $3.14\times 10^{-3}$W

  4. $1.80\times 10^{-3}$W

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Correct Option: A
Explanation:
Here, $r = 6 cm = 6 \times 10^{-2}m,$ 
$N = 20, \omega = 40 \, rad \, s^{-1}$
$B = 2 \times 10^{-2} T, R = 8 \Omega$
Maximum emf induced, $\epsilon \, = \, NAB \omega$
$ = \, N(\pi r^2)B \omega$
$ = 20 \times \pi \times (6 \times 10^{-2})^2 \times 2 \times 10^{-2} \times 40 \, = \, 0.18 V$

Average value of emf induced over a full cycle $\epsilon _{av}\, = \, 0$ 

Maximum value of current in the coil, $ I \, = \, \dfrac{\varepsilon }{R} = \dfrac{0.18}{8} = 0.023 \,A$

Average power dissipated, $ P \, = \, \dfrac{\varepsilon I}{2} \, = \, \dfrac{0.18 \times 0.023}{2} \, = \, 2.07 \times 10^{-3} W$

The axle of a circular wheel of radius R is held horizontally by two identical strings of equal lengths separated by a distance D. The tension in each string is $T _0$. The rim of the wheel carries a total charge $+$Q distributed uniformly on it. The wheel is vertical and is kept in a uniform vertical magnetic field $\vec{B}$. It is now rotated at an angular speed $\omega$. If the string break at a tension of $3T _0/2$, than the maximum possible value of $\omega$ at which the wheel can be rotated without breaking a string is $\dfrac{DT _0}{QBR^2}$.

motional emf physics

  1. True

  2. False

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Correct Option: A

A conductor of $3\ m$ length is moving perpendicular to its length as well as a magnetic field of ${10}^{-3}\ T$ with a speed of ${10}\ m/s$, then the force required to move it with this constant speed is

motional emf physics

  1. $0.3\ N$

  2. $0.9\ N$

  3. Zero

  4. $3 \times {10}^{-3}\ N$

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Correct Option: C
Explanation:

Force on rod moving in a magnetic fileld is given by:

$F=BIL$
Where $I$ is the current in the rod.

According to the question;
$I=0$
Therefore,
$F=0$

Thus,
No force is exerted on the conductor in the magnetic field.
Thus, no external field is required to move the conductor at constant speed.

A horizontal straight conductor (otherwise placed in a closed circuit) along east-west direction falls under gravity. There is:-

motional emf physics

  1. No induced electromotive force along the length

  2. No induced current along the length

  3. An induced current from west to east

  4. None of the above.

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Correct Option: C

In a uniform magnetic field of induction $B$ a wire in the form of semicircle of radius $r$ rotated about the diameter of the circle with angular frequency $\omega$. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is:

motional emf physics

  1. $\dfrac { B\pi { r }^{ 2 }\omega }{ 2R }$

  2. $\dfrac { { \left( B\pi { r }^{ 2 }\omega \right) }^{ 2 } }{ 2R }$

  3. $\dfrac { { \left( B\pi r\omega \right) }^{ 2 } }{ 2R }$

  4. $\dfrac { { \left( B\pi r{ \omega }^{ 2 } \right) }^{ 2 } }{ 8R }$

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Correct Option: D

To produce a field of magnetic $\pi$ tesla at the center of circular loop of diameter 1 m, the current flowing through loop is :

motional emf physics

  1. $5\times 10^6\;A$

  2. $10^7\;A$

  3. $2.5\times 10^6\;A$

  4. $2\times 10^6\;A$

Show answer
Correct Option: C
Explanation:

Magnetic field due to circular loop at center O due to current $i$ is given as,

$B=\dfrac{\mu _0 I}{2r}=\dfrac{4\pi\times 10^{-7} \times I}{2\times 0.5}$
$\pi \times 1=4\pi \times 10^{-7} I$
$I=2.5\times 10^{6} A$

A metal ring of radius $r = 0.5\ m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2\ T$ carries a current $I = 100\ A$. The tension in newtons developed in the ring is

motional emf physics

  1. $100$

  2. $50$

  3. $25$

  4. $10$

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Correct Option: D

A magnet is brought towards a coil (i) speedly (ii) slowly, then the induced electromotive force charge will be respectively

motional emf physics

  1. More in first case / More in first case

  2. More in first case / Equal in both case

  3. Less in first case / More in second case

  4. Less in first case / Equal in both case

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Correct Option: B

An electric charge $+ q$ moves with velocity $\bar{V}=3\hat{i}+4\hat{j}+\hat{k}$, in an electromagnetic field given by $\bar{E}=3\hat{i}+\hat{j}+\hat2{k}$ $\bar{B}=\hat{i}+\hat{j}+\hat3{k}$.The y-component of the force experienced by +q is:

motional emf physics

  1. $-7 q$

  2. $11 q$

  3. $5 q$

  4. $3 q$

Show answer
Correct Option: A
Explanation:

Given that,

Velocity, $v=\left( 3\hat{i}+4\hat{j}+\hat{k} \right)$

Electric field, $E=3\hat{i}+\hat{j}+2\hat{k}$

Magnetic field, $B=\hat{i}+\hat{j}+3\hat{k}$

Force,

$ F=q\left( \text{E+v}\times \text{B} \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+\left( (3\hat{i}+4\hat{j}+\hat{k})\times (\hat{i}+\hat{j}+3\hat{k} \right) \right) $

$ F=q\left( 3\hat{i}+\hat{j}+2\hat{k}+11\hat{i}-8\hat{j}-\hat{k} \right) $

$ F=q\left( 14\hat{i}-7\hat{j}+\hat{k} \right) $

So, y-component of force experienced by +q is

${{F} _{y}}=-7q$

A boat is moving due East in region where the earth's magnetic field is $5.0 \times {10^{ - 5}}\,N{A^{ - 1}}{m^{ - 1}}$ North and horizontal. The boat carries a vertical aerial $2 m$ long. if the speed of the boat is $1.50\,ms - 1,$ the magnitude of the induced emf in the wire of aerial is 

motional emf physics

  1. $0.75 mV$

  2. $0.50 mV$

  3. $40.15 mV$

  4. $1 mv$

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Correct Option: B